hw 20 - absolute extrema section 3.6: problem 8\n(1 point)\nfind the absolute maximum and minimum values of…

hw 20 - absolute extrema section 3.6: problem 8\n(1 point)\nfind the absolute maximum and minimum values of ( f(x)=4 x-9+2 x^{2} ), if any, over the interval ( (-infty,+infty) ).\nabsolute maximum is and it occurs at ( x= )\nabsolute minimum is and it occurs at ( x= )\n(round to three decimal places as needed.)\nnotes: if there is more than one ( x ) value, enter as a comma separated list. enter \none\ in any unused answer box.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 6 attempts remaining.

hw 20 - absolute extrema section 3.6: problem 8\n(1 point)\nfind the absolute maximum and minimum values of ( f(x)=4 x-9+2 x^{2} ), if any, over the interval ( (-infty,+infty) ).\nabsolute maximum is and it occurs at ( x= )\nabsolute minimum is and it occurs at ( x= )\n(round to three decimal places as needed.)\nnotes: if there is more than one ( x ) value, enter as a comma separated list. enter \none\ in any unused answer box.\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 6 attempts remaining.

Answer

Explanation:

Step1: Find the derivative of the function

The function is (f(x)=2x^{2}+4x - 9). Using the power rule ((x^{n})^\prime=nx^{n - 1}), the derivative (f^\prime(x)=(2x^{2}+4x - 9)^\prime=4x + 4).

Step2: Find the critical points

Set (f^\prime(x)=0), so (4x+4 = 0). Solving for (x): [ \begin{align*} 4x+4&=0\ 4x&=-4\ x&=- 1 \end{align*} ]

Step3: Find the second - derivative

Differentiate (f^\prime(x)=4x + 4) to get (f^{\prime\prime}(x)=4). Since (f^{\prime\prime}(x)=4>0), the function (f(x)) has a local minimum at (x =-1).

Step4: Find the value of the function at the critical point

Substitute (x=-1) into (f(x)): [ \begin{align*} f(-1)&=2(-1)^{2}+4(-1)-9\ &=2-4 - 9\ &=-11 \end{align*} ] As (x\to\pm\infty), (y = f(x)=2x^{2}+4x - 9\to+\infty) (because the leading coefficient of the quadratic function (y = 2x^{2}+4x - 9) is (a = 2>0)).

Answer:

absolute maximum is none and it occurs at (x=) none; absolute minimum is (-11) and it occurs at (x=-1)